What is the probability that a normal number will look periodic? Let's take for example $\pi$. The numbe $\pi$ is irrational, hence the decimal pattern is non-periodic. However, could it happen, that by just observing a huge but finite amount of digits of $\pi$, it would "look periodic"?
It is not known, but it is conjectured that $\pi$ is a normal number, i.e. any sequence of digits of length $n$ occurs with density $10^{-n}$. So let's assume this is true, and then ask what is the probability that for some sufficiently large $n$, $\pi$ looks like
$$\pi\approx 3.\underbrace{x_1x_2\cdots x_n}_{\text{first period}}\overbrace{x_1x_2\cdots x_n}^{\text{second period}}\underbrace{\cdots}_{\llap{\text{other}}\,\rlap{\text{digits}}}\;.$$
Of course, this question is not really restricted to $\pi$ or to digits of numbers, but can be asked in the context of normal sequences (joriki pointed out in the answer that for a specific number the probabilit is always one or zero).

Remark. I know that general periodic number can have the form $$0.x_1x_2\cdots x_n\overline{y_1y_2\cdots y_m}$$ with an initial non-periodic sequence $x_1x_2\cdots x_n$. But since in a normal number any finite sequence occures, it is certain that any sequence will also repeat an arbitrary amount of times. The question is specifically about whether such a period can start with the first digit.
 A: One way to answer your question would be: “Since $\pi$ is a fixed number and not a random variable, the probability is either $0$ or $1$; we just don't know what it is.” I suspect that that's not the interpretation you intended for your question, so I'll flesh out the concept of “probabilities” in this context in a different way, by replacing $\pi$ by a number whose decimal digits are randomly uniformly and independently chosen.
Then the probability is not $0$, since the probability for the period $n=1$ to work is $\frac1{10}$.
The probability is also not $1$, since it's bounded above by the sum of the marginal probabilities for all $n$:
$$
\sum_{n=1}^\infty10^{-n}=\frac1{10}\cdot\frac1{1-\frac1{10}}=\frac19\;.
$$
Determining the exact probability seems hard, since one would have to account for all cases with multiple periods. This Java code determines the probability for a period to appear up to finite $n$, with the following results:
\begin{array}{r|l}
1&0.1\\
2&0.109\\
3&0.1099\\
4&0.1099891\\
5&0.109998001\\
6&0.1099988911\\
7&0.109998980101\\
8&0.109998989001028\\
9&0.10999898989102999\\
10&0.1099989899800301008
\end{array}
So the probability is apparently very close to $0.11$, which is the estimate you get by subtracting out the cases where a sequence with a longer period starts with two identical digits:
$$
\frac1{10}+\frac9{10}\sum_{n=2}^\infty10^{-n}=\frac1{10}+\frac9{10}\cdot\frac1{100}\cdot\frac1{1-\frac1{10}}=\frac1{10}+\frac1{100}=0.11\;.
$$
